Question

Bài 1: X = \(\dfrac{b^2+c^2-a^2}{2bc}\) ; Y = \(\dfrac{a^2-\left(b-c\right)^2}{\left(b+c\right)^2-a^2}\)

Tính P = x +y +xy

Giup mk nha các bạn

Answer 1

(x+1)(y+1)=xy+x+y+1 => P=xy+x+y= ( x+1)(y+1)-1

\(\left(x+1\right)=\dfrac{\left(b+c\right)^2-a^2}{2bc}=\dfrac{\left(b+c+a\right)\left(b+c-a\right)}{2bc}\)

\(\left(y+1\right)=\dfrac{a^2-\left(b-c\right)^2+\left(b+c\right)^2-a^2}{\left(b+c+a\right)\left(b+c-a\right)}=\dfrac{4bc}{\left(b+c+a\right)\left(b+c-a\right)}\)

\(\Rightarrow\left(x+1\right)\left(y+1\right)=\dfrac{4bc}{2bc}=2=>xy+x+y=2-1=1\)